From: "Daniel Völkerts" <dvoelkerts@...>
Date: 2004-09-14T04:14:53+09:00
Subject: Re: [webrick] How to extract filename from a request?

James Britt wrote:
> So, the part of the URL that comes after /filter/ is data to be passed 
> to some method?

Yeah.That's the way I'd like to process the file through the servlet.


> If you have /filter/ mapped to a servlet, and call 
> http://localhost/filter/someFilterFile, then, in your servlet, 
> request.path should give you "/filter/someFilterFile"
> 
>   def do_GET(req, res)
>     url_parts = req.path.split( '/' )
>     base = url_parts[0]
>     filter_file_path = url_parts[1]
>   end
> 
> 
> That's one way to grab the data.  (Not the best; no error checking, etc.)

So do I. I extend it with a kind of error correction and it works. 
Thanks in advance.

-- 
Daniel V�lkerts
"Irren ist menschlich, und noch menschlicher ist es,
einem Computer die Schuld zu geben." - U